By Kubota T.

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D. We have (6, 15, 21) = 3. e. We have (−7, 28, −35) = 7. f. 18. We have (66,105,42) = 3, (66,105,70) = 1, (66,105,165) = 3, (66,42,70) = 2, (66,42,165) = 3, (66,70,165) = 1, (105,42,70) = 7, (105,70,165) = 5, and (42,70,165) = 1. Hence there are three sets of mutually relatively prime integers in this set, namely {66, 105, 70} , {66, 70, 105},and {42, 70, 165}. 19. Let d | ai , 1 ≤ i ≤ n. Then clearly dc | cai , 1 ≤ i ≤ n. So dc | (ca1 , ca2 , . . , can ). To see the other direction, note that c | cai for all i, so c | (ca1 , ca2 , .

C. We have 10 is O(n) since 10 ≤ 10n whenever n is a positive integer. d. We have n2 + 1 ≤ 2n2 whenever n is a positive integer. Hence log(n2 + 1) ≤ log(2n2 ) = 2 log n + log 2 ≤ 3n whenever n is a positive integer. It follows that log(n2 + 1) is O(n). e. Note that f. We have (n2 + 1)/(n + 1) < (2n2 /n = 2n whenever n is a positive integer. Hence (n2 + 1)/(n + 1) is O(n). √ n2 + 1 ≤ √ 2n2 ≤ √ 2 · n whenever n is a positive integer. Hence √ n2 + 1 is O(n). Note that for n ≥ 1, 2n4 + 3n3 + 17 ≤ 2n4 + 3n4 + 17n4 = 22n4 .

Then x = x0 + at, y = y0 + bt is a solution for any integer t. Choose t0 so that n − b < y0 + bt0 ≤ n. Then x = x0 + at0 , y = y0 + bt0 is a solution such that (x, y) = 1 and 0 ≤ n − b < y ≤ n. Since x/y is in lowest terms, and y ≤ n, it is a fraction of the nth Farey series. Also a 1 a x = + > , y b by b so that x/y comes later than a/b in the series. If it is not c/d, then it comes later in the series than c/d, and x c dx − cy 1 − = ≥ , y d dy dy and c a bc − ad 1 − = ≥ . d b bd bd Hence 1 bx − ay x a 1 1 b+y n 1 = = − ≥ + = > ≥ , by by y b dy bd bdy bdy by which is a contradiction.